poj 1050 To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 45906		Accepted: 24276

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:

9 2
-4 1
-1 8
and has a sum of 15.

Input

The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output

Output the sum of the maximal sub-rectangle.

Sample Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

Sample Output

15

Source

Greater New York 2001

翻译：

总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

已知矩阵的大小定义为矩阵中所有元素的和。给定一个矩阵，你的任务是找到最大的非空(大小至少是1 * 1)子矩阵。

比如，如下4 * 4的矩阵

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

的最大子矩阵是

9 2
-4 1
-1 8

这个子矩阵的大小是15。

输入

输入是一个N * N的矩阵。输入的第一行给出N (0 < N <= 100)。再后面的若干行中，依次（首先从左到右给出第一行的N个整数，再从左到右给出第二行的N个整数……）给出矩阵中的N²个整数，整数之间由空白字符分隔（空格或者空行）。已知矩阵中整数的范围都在[-127, 127]。

输出

输出最大子矩阵的大小。

样例输入

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

样例输出

15

第一种

/*准备一个数组F[i,j]来存到第i,j格时的矩阵和(类似于前缀和)
对于一个子矩阵[x1,y1,x2,y2]//x1,x2代表左上角和右下角的横坐标
有 S[x1,y1,x2,y2] = f[x2,y2] - f[x1-1,y2] - f[x2,y1-1] + f[x1,y1];*/
#include<cstdio>
#include<iostream>
using namespace std;
#define N 101
int ans=-0x7f,n,a[N][N],f[N][N];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            scanf("%d",&a[i][j]);
            f[i][j]=f[i-1][j]+f[i][j-1]-f[i-1][j-1]+a[i][j];
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            for(int k=0;k+i<=n;k++)
                for(int l=0;l+j<=n;l++){
                    int xx=i+k,yy=j+l;
                    ans=max(ans,f[xx][yy]-f[i-1][yy]-f[xx][j-1]+f[i-1][j-1]);
                }
    printf("%d
",ans);
    return 0;
}

第二种

#include<cstdio>
#include<iostream>
using namespace std;
#define N 101
int a[N][N],n,ans=0;
int main(){
    scanf("%d",&n);
    for(int i=1,x;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%d",&x),a[i][j]=a[i-1][j]+x;//列前缀和
    for(int i=1;i<=n;i++)
        for(int j=i;j<=n;j++){
            int tmp=0;
            for(int k=1;k<=n;k++){
                int num=a[j][k]-a[i-1][k];
                if(tmp>0) tmp+=num;
                else tmp=num;
                ans=max(ans,tmp);
            }
        }
    printf("%d
",ans);
    return 0;
}